‌α+β+γ=a ‌αβ+βγ+γα=a ‌αβγ=1 ‌⇒α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα) ‌‌‌⇒α2β2+β2γ2+γ2α2=(αβ+βγ+γα)2−2αβγ(α+β+γ) ‌‌‌=a2−2a ‌‌‌∴α2β2γ2=(αβγ)2=12=1 Now, equate the coefficients of the original and new equations. a2−2a=a a2−2a=a 1=1 Solve for a ‌a2−2a=a ‌⇒‌‌a2−3a=0 ‌⇒‌‌a(a−3)=0 ‌∵‌‌a≠0∴a=3
